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Question Number 202127 by Calculusboy last updated on 21/Dec/23

Answered by qaz last updated on 21/Dec/23

$${\int }_{\pi /6}^{\pi /3}\frac{dx}{1+{\tan }^{\pi }x}={\int }_{\pi /6}^{\pi /3}\frac{dx}{1+\cot {}^{\pi }x}=\frac{1}{2}{\int }_{\pi /6}^{\pi /3}(\frac{1}{1+{\tan }^{\pi }x}+\frac{1}{1+\cot {}^{\pi }x})dx$$$$=\frac{1}{2}{\int }_{\pi /6}^{\pi /3}dx=\frac{\pi }{4}$$

Commented by Calculusboy last updated on 22/Dec/23

$$\mathit{n}\mathit{i}\mathit{c}\mathit{e}\mathit{s}\mathit{o}\mathit{l}\mathit{u}\mathit{t}\mathit{i}\mathit{o}\mathit{n}\mathit{s}\mathit{i}\mathit{r}$$

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